The displacement (in meter) of a particle moving along the x-axis is given $x = 18 t + 5 t^{2} .$ Calculate
(A) The instantaneous velocity at t = 2 sec
(B) Average velocity between t = 2 sec and t = 3 sec

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Solution

$\begin{matrix} x = 5 t^{2} + 18 t (1) v = \frac{d x}{d t} = 10 t + 18 \end{matrix}$
$a t t = 2 s, v = 10 \times 2 + 18 = 38 m / s$
$\begin{matrix} (2) A v e r a g e, v e l o c i t y = \frac{d i s p l a c e m e n t}{t i m e} a t t = 2, x = 5 (2)^{2} + 18 \times 2 = 56 m t = 3, x = 5 (3)^{2} + 18 \times 3 = 99 m A v e r a g e v e l o c i t y = \frac{99 - 56}{3 - 2} = 43 m / s \end{matrix}$

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